In: Chemistry
1) Zinc reacts with hydrochloric acid according to the reaction equation.How many milliliters of 4.00 M HCl(aq) are required to react with 5.85 g of an ore containing 44.0% Zn(s) by mass?
2) The amount of I3–(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32–(aq) (thiosulfate ion). The determination is based on the net ionic equation.
2SO2O3(2-)(aq) +I3(-)(aq) --->>> S4O6(2-)(aq) + 3I(-)(aq)
Given that it requires 25.6 mL of 0.460 M Na2S2O3(aq) to titrate a 10.0-mL sample of I3–(aq), calculate the molarity of I3–(aq) in the solution.
1) Zinc reacts with hydrochloric acid according to the reaction equation.How many milliliters of 4.00 M HCl(aq) are required to react with 5.85 g of an ore containing 44.0% Zn(s) by mass?
Solution :-
Balanced reaction equation
Zn + 2HCl ----- > ZnCl2 + H2
Lets calculate the mass of the Zn using the percent
5.85 g * 44.0 % Zn / 100 % = 2.574 g Zn
Now lets calculate the moles of the Zn
Moles of Zn = mass of Zn / molar mass of Zn
= 2.574 g Zn / 65.39 g per mol
= 0.03936 mol Zn
Now lets calculate the moles of the HCl using the mole ratio of the Zn and HCl
0.03936 mol Zn * 2 mol HCl / 1 mol Zn = 0.07872 mol HCl
Now lets calculate the volume of the HCl
Volume in liter = moles / molarity
= 0.07872 mol / 4.00 mol per L
= 0.0197 L
0.0197 L * 1000 ml / 1 L= 19.7 ml
So the volume of HCl needed = 19.7 ml
2) The amount of I3–(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32–(aq) (thiosulfate ion). The determination is based on the net ionic equation.
2SO2O3(2-)(aq) +I3(-)(aq) --->>> S4O6(2-)(aq) + 3I(-)(aq)
Given that it requires 25.6 mL of 0.460 M Na2S2O3(aq) to titrate a 10.0-mL sample of I3–(aq), calculate the molarity of I3–(aq) in the solution.
Solution :-
Lets first calculaete the moles of the Na2S2O3
Moles = molarity * volume in liter
= 0.460 mol per L * 0.0256 L
= 0.01178 mol Na2S2O3
Now lets calculate the moles of the I3^-
0.01178 mol S2O3^2- * 1 mol I3^- / 2 mol S2O3^2- = 0.00589 mol I3^-
Now lets calculate the molarity of the I3^-
Molarity = moles / volume in liter
10 ml * 1 L / 1000 ml = 0.010 L
So molarity = 0.00589 mol I3^- / 0.010 L
= 0.589 M
So the molarity of the I3^- = 0.589 M